**The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has both U and V as descendants.**

**A binary search tree (BST) is recursively defined as a binary tree which has the following properties:**

**The left subtree of a node contains only nodes with keys less than the node’s key.**

**The right subtree of a node contains only nodes with keys greater than or equal to the node’s key.**

**Both the left and right subtrees must also be binary search trees.**

**Given any two nodes in a BST, you are supposed to find their LCA.**

**Input Specification:**

**Each input file contains one test case. For each case, the first line gives two positive integers: M (<= 1000), the number of pairs of nodes to be tested; and N (<= 10000), the number of keys in the BST, respectively. In the second line, N distinct integers are given as the preorder traversal sequence of the BST. Then M lines follow, each contains a pair of integer keys U and V. All the keys are in the range of int.**

**Output Specification:**

**For each given pair of U and V, print in a line “LCA of U and V is A.” if the LCA is found and A is the key. But if A is one of U and V, print “X is an ancestor of Y.” where X is A and Y is the other node. If U or V is not found in the BST, print in a line “ERROR: U is not found.” or “ERROR: V is not found.” or “ERROR: U and V are not found.”.**

**Sample Input:**

**6 8**

**6 3 1 2 5 4 8 7**

**2 5**

**8 7**

**1 9**

**12 -3**

**0 8**

**99 99**

**Sample Output:**

**LCA of 2 and 5 is 3.**

**8 is an ancestor of 7.**

**ERROR: 9 is not found.**

**ERROR: 12 and -3 are not found.**

**ERROR: 0 is not found.**

**ERROR: 99 and 99 are not found.**

**题目大意：给出一棵二叉搜索树的前序遍历，问结点u和v的共同最低祖先是谁～**

**分析：map<int, bool> mp用来标记树中所有出现过的结点，遍历一遍pre数组，将当前结点标记为a，如果u和v分别在a的左、右，或者u、v其中一个就是当前a，即(a >= u && a <= v) || (a >= v && a <= u)，说明找到了这个共同最低祖先a，退出当前循环～最后根据要求输出结果即可～**

**PS：30分的题目30行代码解决，1行1分，惊不惊喜？意不意外？（真的是水题啊…）**

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#include <iostream> #include <vector> #include <map> using namespace std; map<int, bool> mp; int main() { int m, n, u, v, a; scanf("%d %d", &m, &n); vector<int> pre(n); for (int i = 0; i < n; i++) { scanf("%d", &pre[i]); mp[pre[i]] = true; } for (int i = 0; i < m; i++) { scanf("%d %d", &u, &v); for(int j = 0; j < n; j++) { a = pre[j]; if ((a >= u && a <= v) || (a >= v && a <= u)) break; } if (mp[u] == false && mp[v] == false) printf("ERROR: %d and %d are not found.\n", u, v); else if (mp[u] == false || mp[v] == false) printf("ERROR: %d is not found.\n", mp[u] == false ? u : v); else if (a == u || a == v) printf("%d is an ancestor of %d.\n", a, a == u ? v : u); else printf("LCA of %d and %d is %d.\n", u, v, a); } return 0; } |